Answer:
The number of moles of oxygen present in the crew cabin at any given time is 615.309 moles
Explanation:
The given parameters are;
Volume of the crew cabin = 74,000 L
Pressure of the crew cabin = 1.00 atm
Percentage of nitrogen in the mixture of gases in the cabin = 80%
Percentage of oxygen in the mixture of gases in the cabin = 20%
Temperature of the cabin = 20°C = 293.15 K
Therefore, volume of oxygen in the crew cabin = 20% of 74,000 L
Hence, volume of oxygen in the crew cabin = ![\frac{20}{100} \times 74,000 \, L = 14,800 \, L](https://tex.z-dn.net/?f=%5Cfrac%7B20%7D%7B100%7D%20%5Ctimes%2074%2C000%20%5C%2C%20L%20%3D%2014%2C800%20%5C%2C%20L)
From the universal gas equation, we have;
![n = \frac{P \times V}{R \times T}](https://tex.z-dn.net/?f=n%20%3D%20%5Cfrac%7BP%20%5Ctimes%20V%7D%7BR%20%20%5Ctimes%20%20T%7D)
Where:
n = Number of moles of oxygen
P = Pressure = 1.00 atm
V = Volume of oxygen = 14,800 L
T = Temperature = 293.15 K
R = Universal Gas Constant = 0.08205 L·atm/(mol·K)
Plugging in the values, we have;
![n = \frac{1 \times 14,800 }{0.08205 \times 293.15 } = 615.309 \, moles](https://tex.z-dn.net/?f=n%20%3D%20%5Cfrac%7B1%20%5Ctimes%2014%2C800%20%7D%7B0.08205%20%20%20%5Ctimes%20%20293.15%20%7D%20%3D%20615.309%20%5C%2C%20moles)
The number of moles of oxygen present in the crew cabin at any given time = 615.309 moles.